The lamda-nucleolus for non-transferable
utility games and Shapley procedure
Chih Chang
Department of Mathematics
National Tsing Hua University
Hsinchu, Taiwan
E-mail: [email protected]
Peng-An Chen
Department of Mathematics
National Taitung University
Taitung, Taiwan
E-mail: [email protected]
July 23, 2013
Abstract
We …rst scrutinize two extensions of the nucleolus for NTU games
by Kalai(1975) and Nakayama(1983) ; respectively. We see that some
nice things disappear during the process of extension. To de…ne the
-nucleolus, the -excess function is de…ned and is axiomatized. It is
nice to see that the -nucleolus satis…es the coincidence property and
preserves many nice properties of the nucleolus for TU games. Shapley
procedure proposed by Shapley(1969) is used to extend Shapley value
to NTU games. It is known that the procedure can be applied to a
The authors wish to express their gratitude to Mamoru Kaneko for many helpful
comments and discussions on both substantive matters and exposition which much improve
the paper.
1
variety of solutions for TU games to obtain their corresponding extensions for NTU games. In particular, the -nucleolus is the nucleolus
extension derived via the procedure. If one tries to extend a solution
for TU games to NTU cases, the one derived via Shapley procedure is
recommended.
Keywords: TU games, the nucleolus, NTU games, the -excess
function, the -nucleolus, continuity, symmetry, upper hemi-continuity,
Shapley procedure
JEL Classi…cation C71
1
Introduction
To de…ne a solution for games without transferable utilities (NTU games),
one usually employs the ideas of solutions for games with transferable utilities (TU games). In this paper, we study several possible extensions of the
nucleolus for NTU games. To see whether an extension is justi…ed, there are
two criteria to be considered:
(A) The coincidence property: The extension coincides with its original
version on TU games.
(B) The preservation criterion: The extension preserves nice properties
of its original version.
It is known that the nucleolus for TU games satis…es symmetry, continuity, and the core property which means that a solution is contained in the
core if the core is non-empty, please see Schmeidler(1969). For criterion (B) ;
we specify these three appealing properties to be studied.
We …rst scrutinize two existing extensions proposed by Kalai(1975) and
Nakayama(1983) ; respectively; and show that each one does not satisfy all
the three properties. Furthermore, an impossibility result is presented to
illustrate that there is no non-empty solution for NTU games satisfying the
three properties. It means that some nice things disappear during the process
of extension.
Among several possible extensions that we study, the -nucleolus is recommended because it satis…es (A) and (B) to a large extend. Besides, it is not
di¢ cult to provide an axiomatization of the -nucleolus to justify whether
it is game-theoretically sound. Indeed, the idea that Aumann(1985) used to
axiomatize the NTU value for NTU games can be employed. According to
these, the -nucleolus seems to be a nice extension of the nucleolus.
2
Moreover, the -excess function which is used to derive the -nucleolus
for NTU games is axiomatized as well. It is the unique function satisfying
excess preservation, excess additivity, and excess invariance.
Shapley procedure proposed by Shapley(1969) is used to extend the Shapley value to NTU games. It is known that the procedure can be applied to a
variety of solutions for TU games. In particular, the NTU value and the nucleolus are extensions of the Shapley value and the nucleolus, respectively,
derived via the procedure. We see that each of these extensions shares nice
properties of its original version, too. It indicates that Shapley procedure
provides a good extension if one tries to extend a solution for TU games to
NTU cases.
The paper is organized as follows. Section 2 is devoted to specify notations used, game models, and properties of solutions. In section 3, we
scrutinize properties of both Kalai’s extension and Nakayama’s extension: A
non-continuity Theorem is presented. An impossibility Theorem is proposed
in section 4. In section 5, the -excess function is axiomatized and the nucleolus is de…ned. We then study the issue of axiomatization in section
6.
2
De…nitions and notations
In this section, we introduce some basic de…nitions and concepts to be discussed in this paper, in particular, TU games, NTU games, and two wellknown solution concepts: the core and the nucleolus. Several properties for
solutions are de…ned, too.
Let N = f1; 2; ; ng be the player set: We call a non-empty subset S of
N a coalition and denote P the set of all coalitions: Points in RN are called
payo¤ vectors. Given a payo¤ vector x and S 2 P; we denote x (S) = i2S xi
and xS the restriction of x on RS ; where
RS = x 2 RN : xi = 0 if i 2
=S :
Denote GS (fxg) = z 2 RS j zi xi 8 i 2 S and GS (A) = [a2A GS (fag) ;
where A RN :
Let
RS+ = x 2 RS : xi 0 if i 2 S ;
RS++ = x 2 RS : xi > 0 if i 2 S ;
3
and
N
= x 2 RN
+ : x (N ) = 1 :
Let D RN : Denote int (D) and ri (D) to be the interior and the relative
interior of the set D; respectively. Denote @D and r@(D) to be the boundary
and the relative boundary of D; respectively: If D has an interior point, it is
known that int (D) = ri (D) and @D = r@(D): For de…nitions of these terms,
please see Rockafellar(1970) : If D RS ; we say that D is S-comprehensive
if x 2 D; y 2 RS ; and xi yi 8i 2 S; then y 2 D:
A TU game is a pair (N;
P v) ; where v assigns to each coalition S a real
number v (S) and v (S)
i2S v (fig) 8S 2 P: We call v0 2 GN the zero
game if v0 (S) = 0 8S 2 P. Denote GN to be the set of all TU games
with the player set N: For convenience, we use the small letter v to represent
(N; v) if no confusion arises:
Given a game v 2 GN ; the pre-imputation set of v is
I (v) = fx 2 RN : x (N ) = v(N ) g;
and the imputation set of v is
I (v) = fx 2 RN : x 2 I (v) and xi
v(fig) for all i 2 N g:
Each element in I (v) is called an imputation.
The core of the game v 2 GN is de…ned to be
C (v) = fx 2 I (v) j x (S)
v (S) for all S 2 Pg :
The idea to de…ne the nucleolus for TU games is to minimize “total
dissatisfaction” among the elements in I (v) in the sense of lexicographic
order. To introduce the de…nition of the nucleolus for TU games, we need
various concepts.
The excess e is a real-valued function de…ned on RN P GN such that,
given x 2 RN ; S 2 P, and v 2 GN ;
e (x; S; v) = v (S)
x (S) :
The interpretation of the excess is to measure “dissatisfaction” of coalition
S at x: We reshu- e the set fe (x; S; v) : S 2 Pg in the descending way, i.e.,
e (x; Sk ; v) e (x; Sk+1 ; v) for all k = 1; 2; ; 2n 2: The resulting vector is
(x; v; e) = (e (x; S1 ; v) ; e (x; S2 ; v) ;
4
; e (x; S2n 1 ; v)) :
(1)
In a word, (x; v; e) orders the “dissatisfactions”of coalitions at x; the highest “dissatisfaction”…rst, the second-highest “dissatisfaction”second, and so
forth.
If x; y 2 RN ; we say that (x; v; e) is lexicographically less than (y; v; e) ;
denoted by (x; v; e) (lex) (y; v; e) ; if and only if there is an integer k,
where 1
k
2n 1, such that the …rst k-1 components of (x; v; e)
and (y; v; e) are equal and the kth component of (x; v; e) is less than
the kth component of (y; v; e). We denote (x; v; e)
(lex) (y; v; e) if
(x; v; e) (lex) (y; v; e) or (x; v; e) = (y; v; e) :
The nucleolus of v 2 GN is de…ned to be
N (v; e) = f x 2 I (v) j (x; v; e)
(lex)
(y; v; e) 8y 2 I (v)g:
Schmeidler(1969) proved that N (v; e) is single-valued if I (v) 6= ;:
Before going to de…ne NTU games, we need more notations. Given x; y 2
N
R ; we denote x
y if xi < yi 8i 2 N ; x < y if x 6= y and xi
yi
8i 2 N ; and x y if xi yi 8i 2 N: Set x y = x1 y1 + x2 y2 +
+ xn yn ;
x y = (x1 y1 ; x2 y2 ; :::; xn yn ) ; and xy to be the line segment connecting x and
y:
An NTU game is a pair (N; V ), where V assigns to each coalition S a
proper subset of RS satisfying: for all S 2 P;
(A1) V (S) is closed, non-empty, and S-comprehensive,
(A2) IR (V (S)) is non-empty and bounded,
where
IR (V (S)) = fx 2 V (S) j xi
yi 8 y 2 V (fig) and 8 i 2 Sg :
In other words, IR (V (S)) is the individually rational part of V (S) : Denote I (V ) = @V (N ) \ IR (V (N )) : We call the payo¤ vector in I (V ) an
imputation as well.
Denote GN to be the set of NTU games with the player set N: For convenience, (N; V ) is simply represented by the capital letter V if no confusion
arises.
The core of a game V 2 GN is de…ned to be
C (V ) = f z 2 V (N ) j 9/ y 2 V (S) with yS
5
zS 8S 2 Pg:
Given v 2 GN and V 2 GN ; we call v and V corresponding to each other
if, for every S 2 P;
V (S) = x 2 RS : x (S)
v(S) :
If a game v 2 GN corresponds to V 2 GN ; the game v can be viewed as a
game in GN ; and vice versa. If V0 corresponds to v0 ; then V0 2 GN and V0 is
called the zero game, too. We observe that I(V ) = I (v) and C (V ) = C (v)
if v and V correspond to each other.
V 2 GN is called a quasi hyperplane game if there are h 2 RN
++ and a real
constant c such that
V (N ) = f x 2 RN : h x
cg.
The last part of the section is to de…ne three properties of a solution that
GN is a mapping
are relevant to the paper. A solution de…ned on G0N
satisfying (v) I (v) for all v 2 G0N .
Given any v 2 G0N with C (v) 6= ;; we say that satis…es the core property
if (v) C (v) :
Denote to be a permutation on N: is called a symmetry of v 2 G0N if
v ( S) = v (S) 8 S 2 P. We say that is symmetric if (v) = ( (v)) for
all v 2 G0N and every symmetry of v:
Let be a single-valued solution. Given a sequence of games fvn g1
n=1
G0N and vn ! v 2 G0N ; is continuous if (vn ) ! (v) as n ! 1:
Next, we extend these three properties to NTU games. A solution
0
GN is a mapping satisfying (V ) @V (N ) : We say that
de…ned on GN
0
with C (V ) 6= ;; then (V ) C (V ) :
satis…es the core property if V 2 GN
Let be a permutation on N and S 2 P. Given x 2 RS and V 2
0
GN ; we denote (x) 2 R (S) ; where ( (x)) (i) = xi 8i 2 S; and V (S) =
0
f (x) : x 2 V (S)g : is a symmetry of V 2 GN
if V ( S) = V (S) 8S 2 P.
0
We say that is symmetric if (V ) = ( (V )) for every V 2 GN
and every
symmetry of V:
Given S 2 P, we denote F to be the collection of non-empty closed
subsets of RS and hS : F F ! R [ f 1; +1g an extended Hausdor¤
metric, please see Aliprantis and Border(2006) for details. The distance of
two games is de…ned to be: if V; W 2 GN ;
(V; W ) = max hS (V (S) ; W (S)) :
S2P
6
0
0
Let fVn g1
n=1 be a sequence of games in GN and Vn ! V 2 GN in the
topology induced by the extended Hausdor¤ metric. The solution is upper
0
hemi-continuous (u.h.c.) in GN
if xn 2 (Vn ) 8 n and xn ! x 2 V (N ) as
n ! 1, then x 2 (V ).
3
Two extensions of the nucleolus
Kalai(1975) and Nakayama(1983) each provided an extension of the nucleolus
for NTU games. In this section, we scrutinize whether these two extensions
satisfy the two criteria: (A) The coincidence property, and (B) The preservation criterion. For (B), we will study whether symmetry, continuity, and the
core property are preserved. Some of these have been studied. Both extensions satisfy the core property, please see Kalai(1975) and Nakayama(1983).
Klauke(2002) pointed out that Kalai’s extension does not satisfy the coincidence property. It is not di¢ cult to see whether these two extensions satisfy
symmetry. The main purpose of the section is to show that both extensions
are not u.h.c.
We call a real-valued function E de…ned on RN
P
GN an excess
function for NTU games. Following the same idea that we de…ne N (v; e) for
TU games, the nucleolus for NTU game V derived by E is de…ned to be
N (V; E)
= f x 2 IR(V (N )) j (x; V; E)
(lex) (y; V; E) 8 y 2 IR(V (N ))g,
(2)
where (x; V; E) is the same as (1) except that the TU game v and the excess
e are replaced by V and E; respectively. Note that N (V; E) is not necessarily
single-valued. Hence, instead of continuity, we will see whether it is u.h.c.
For TU games, we know that the excess e is continuous in x 2 RN and
v 2 GN ; and satis…es, for every S 2 P,
8
< = 0; if x (S) = v (S) ;
e (x; S; v) > 0; if x (S) < v (S) ;
:
< 0; if x (S) > v (S) :
It is also known that N (v; e) is continuous.
However, these three phenomena cannot be extended to NTU games altogether.
7
Theorem 1 (Non-continuity) There is no excess function E satisfying
the following three conditions simultaneously:
(i) E is continuous in x 2 RN and V 2 GN , 8
< = 0; if xS 2 r@V (S) ;
(ii) If x 2 IR (V (N )) and S 2 P; E (x; S; V ) > 0; if xS 2 ri(V (S));
:
< 0; if xS 2
= V (S) ;
(iii) N (V; E) is u.h.c.
The idea to prove the fact is the following: We …rst propose a sequence
of games fVn g1
n=1 which converges to V: Then we show that N (Vn ; E) 9
N (V; E) as n ! 1 if E satis…es conditions (i) and (ii) :
Example 2 Let N = f1; 2; 3g and k be a large enough positive integer. Set
A = (1; 1; 0), An = 1 + k+n1 2 ; 1 + k+n1 2 ; 0 ,
B = (1; 0; 1), Bn = 1 + k+n1 2 ; 0; 1 + k+n1 2 ,
C = Cn = (0; 1; 1) ;
D = (0; 0; 1), Dn = 0; 0; 1 + k+n1 2 ,
E = 0; 1 k1 ; 1 k1 , En = 0; 1 k1 1 n1 ; 1
F = (0; 1; 0) ; and Fn = 0; 1 + k+n1 2 ; 0 .
1
k
1
1
n
,
Consider V 2 GN given by
V (fig) = Gfig (f(0; 0; 0)g) for i = 1; 2; 3,
V (f1; 2g) = Gf1;2g (fAg), V (f1; 3g) = Gf1;3g (fBg),
V (f2; 3g) = Gf2;3g (DE [ EF ), and
V (N ) = GN
x 2 RN j x 1 + x 2 + x 3
f1;2;3g
2 \ R+
.
For arbitrary positive integer n, let Vn be de…ned as
Vn (fig) = Gfig (f(0; 0; 0)g) for i = 1; 2; 3,
Vn (f1; 2g) = Gf1;2g (fAn g), Vn (f1; 3g) = Gf1;3g (fBn g),
Vn (f2; 3g) = Gf2;3g (Dn En [ En Fn ), and
f1;2;3g
Vn (N ) = GN (the hyperplane determined by An ; Bn ; and C) \ R+
= GN
x 2 RN j
k+n 3
x
k+n 1 1
+ x2 + x3
f1;2;3g
2 \ R+
Since An ! A, Bn ! B, Dn ! D, En ! E, and Fn ! F as n ! 1; we
have Vn ! V in the topology induced by extended Hausdor¤ metric. Please
see Figure 1.
8
3
(0,0,2)
V0
Dn
Bn
C
D
En
E
B
F
O
Fn
(0,2,0)
2
A
An
(2,0,0)
(2(k+ n -1)/(k+n -3),0,0)
Vn
1
Claim: N (V; E) satis…es the core property if E satis…es the assumption
(ii).
If y 2 IR(V (N )) C (V ) ; there exist a S 2 P and a y 0 2 V (S) such
yS : That is, yS 2 ri(V (S)): We have E(y; S; V ) > 0 by (ii). If
that yS0
x 2 C (V ) 6= ;; then xS 2
= ri (V (S)) ; and hence, E(x; S; V )
0 8S 2 P
by (ii) again. We obtain that (x; V; E) (lex) (y; V; E): N (V; E) satis…es
the core property:
We see that C (V ) = fA; B; Cg and C (Vn ) = fAn ; Bn g 6= ; for all
n. By Claim, N (Vn ; E)
C (Vn ) ! fA; Bg as n ! 1. To show that
N (Vn ; E) 9 N (V; E) ; it is enough to prove that N (V; E) = fCg :
E(A; S; V ) = 0 if S 2 ff3g ; f1; 2g ; f1; 3g ; f2; 3g ; f1; 2; 3gg and E(A; S; V )
< 0 if S 2 ff1g ; f2gg by (ii) : Hence,
(A; V; E) = (0; 0; 0; 0; 0; a1 ; a2 ) ; where a2
a1 > 0:
Similarly, we derive that
(B; V; E) = (0; 0; 0; 0; 0; a3 ; a4 ) ; where a4
a3 > 0;
and
(C; V; E) = (0; 0; 0; 0; a5 ; a6 ; a7 ) ; where a7
a6
a5 > 0:
(C; V; E) is strictly less than both (A; V; E) and (B; V; E) in lexicographical order: We obtain that N (V; E) = fCg :
V and fVn g1
3. Indeed,
n=1 can be viewed as n-person games, where n
let players other than N = f1; 2; 3g be dummy in additive sense.
9
Next, we study whether Kalai’s extension of the nucleolus for NTU games
satis…es u.h.c. Kalai(1975) required an excess function E satisfying the following four additional conditions: Given S 2 P and V 2 GN ;
(a) If x; y 2 RN and xS = yS , then E (x; S; V ) = E (y; S; V ) :
(b) If x; y 2 RN and xS
yS , then E (x; S; V ) > E (y; S; V ) :
N
(c) If x 2 R and xS 2 r@V (S), then E (x; S; V ) = 0.
(d) E is continuous in V and x 2 RN .
Then he proposed a speci…c excess function E satisfying (a) - (d) and
showed that N (V; E) is not u.h.c.
We would like to ask further. Is there an excess function E satisfying
(a) - (d) such that N (V; E) is u.h.c.? The answer is no. Indeed, E satis…es
conditions (i) and (ii) of non-continuity Theorem due to (d) ; and (b) and
(c) ; respectively. Hence, N (V; E) is not u.h.c. for every excess function E
satisfying (a) - (d) by non-continuity Theorem.
The last part of the section is to show that Nakayama’s extension of the
nucleolus is not u.h.c. either. For self-su¢ ciency of the paper, we recall the
idea of the extension brie‡y.
Given a w 2 N ; S 2 P, and V 2 GN ; the payo¤ vector h (w; S) wS 2
r@V (S) is assigned by w to the coalition S; where
h (w; S) = max fh : h wS 2 V (S)g :
(3)
We see that w assigns h (w; N ) w 2 @V (N ) to the coalition N:
The idea to de…ne the extension of the nucleolus is to search a vector
of weights w such that the payo¤ vector h (w; N ) w assigned by w to N
satis…es the following two conditions: (i) it is in I (V ) ; and (ii) it is the
smallest element in a sense of lexicographic order.
To assure h (w; N ) w 2 I (V ) ; we concern only those vectors of weights
in the set
(V ) = w 2
N
: h (w; N ) wi
yi 8 y 2 V (fig) and 8i 2 N :
Indeed, each w 2 (V ) assigns h (w; N ) w 2 I (V ) to N:
The real-valued function eN which plays the role of the excess is de…ned
on (V ) P GN such that, given w 2 (V ) ; S 2 P, and V 2 GN ;
eN (w; S; V ) = (h (w; S) wS ) (S)
(h (w; N ) w) (S) :
That is, eN measures the di¤erence between the sum of the payo¤s assigned
by w to the coalition S and the sum of the payo¤s of players in S assigned
by w to N:
10
Following (1) ; we let
(w; V; eN ) = (eN (w; S1 ; V ) ; eN (w; S2 ; V );
; eN (w; S2n 1 ; V ));
where eN (w; Sk ; V ) eN (w; Sk+1 ; V ) for all k = 1; 2; ; 2n 2:
Nakayama’s extension of the nucleolus, denoted NN (V; eN ) ; consists of
the imputations h (w; N ) w 2 I (V ) ; where w 2 (V ) such that (w; V; eN )
(lex) (w0 ; V; eN ) 8w0 2 (V ) :
In order to employ non-continuity Theorem to show that NN (V; eN ) is
not u.h.c., we will de…ne an excess function EN on RN P GN which is an
extension of eN ; that is, given S 2 P and V 2 GN ; EN (x; S; V ) = eN (x; S; V )
for every x 2 I (V ) : Then the result that NN (V; eN ) is not u.h.c. can be
derived by showing that N (V; EN ) = NN (V; eN ) and N (V; EN ) is not u.h.c.
It will be convenient to divide the argument into three steps. First, we
reformulate the domain of the function eN to be I (V ) P GN instead of
(V ) P GN :
Theorem 3 Let V 2 GN : x 2 I (V ) if and only if there is a w 2
that x = h (w; N ) w: Moreover, x (N ) = h (w; N ) :
(V ) such
x
= w: Then w 2 (V ) and x = x (N ) w:
Proof. Let x 2 I (V ) : Denote x(N
)
It implies x (N ) = h (w; N ) by (3). It is obvious to observe the converse.
We see that every w 2 (V ) corresponds to an x 2 I (V ) ; and vice versa.
Hence, we denote the element in I (V ) corresponding to w 2 (V ) to be xw ;
that is,
xw = h (w; N ) w 2 I (V ) :
From now on, we can view the domain of eN as I (V )
eN (w; S; V ) can be rewritten as
eN (xw ; S; V ) = (h (w; S) wS ) (S)
P
GN : Hence,
xw (S)
and (w; V; eN ) can be replaced by (xw ; V; eN ) : Then
NN (V; eN ) = fxw 2 I (V ) j (xw ; V; eN )
(lex) (xw0 ; V; eN ) 8 xw0 2 I (V )g:
Here we see a by-product that NN (V; eN ) satis…es the coincidence property,
that is, NN (V; eN ) = N (v; e) if V corresponds to the TU game v.
Second, we de…ne the excess function EN ; an extension of eN ; which is
required to be continuous in x and V and to satisfy
EN (x; S; V ) = (h (w; S) wS ) (S)
11
x (S) ;
(4)
x
2 (V ) : We see that EN (x; S; V ) = eN (x; S; V ) if x 2
where w = x(N
)
I (V ) : Hence, EN is an extension of eN :
Lemma 4 Given V 2 GN ; (i) N (V; EN ) = NN (V; eN ); and (ii) N (V; EN )
satis…es the core property
Proof. If y 2 IR (V (N )) I (V ) ; then there is a y 0 2 I (V ) such that
0
y
y < y 0 and y0 y(N ) = y(N
2
(V ) : Then EN (y; S; V )
EN (y 0 ; S; V ) 8
)
S 2 P and EN (y; S; V ) > EN (y 0 ; S; V ) for at least one S 2 P by (4) : Hence,
(y 0 ; V; EN )
(lex) (y; V; EN ): It implies that y 2
= N (V; EN ) : Therefore,
N (V; EN ) must be in I (V ) : Since EN (x; S; V ) = eN (x; S; V ) if x 2 I (V )
and S 2 P, (i) is obtained.
(ii) can be derived by (i) and the fact that N (V; eN ) satis…es the core
property.
Third, we show that N (V; EN ) is not u.h.c. It remains to check that EN
satis…es assumption (ii) of non-continuity Theorem. Given x 2 IR (V (N )) ;
x
we denote w = x(N
; and hence, xS = x (N ) wS for every S 2 P. It implies
)
that
8
< = h (w; S) ; if xS 2 r@V (S) ;
x (N ) < h (w; S) ; if xS 2 ri(V (S));
:
> h (w; S) ; if xS 2
= V (S) ;
and hence, by (4) ;
8
< = 0; if xS 2 r@V (S) ;
EN (x; S; V ) > 0; if xS 2 ri(V (S));
:
< 0; if xS 2
= V (S) :
We have that N (V; EN ) is not u.h.c. by non-continuity Theorem. Then
the desired result that NN (V; eN ) is not u.h.c. is derived by the fact that
N (V; EN ) = NN (V; eN ) of Lemma 4.
Remark 5 Although Nakayama requires a game V satisfying supfyi : y 2
V (fig)g > 0; non-continuity Theorem still holds in the setting because we may
shift the origin so that games considered in Example 2 satisfy the requirement.
4
An impossibility result for a solution
It is known that the nucleolus for TU games satis…es symmetry, continuity,
and the core property. Is there a solution for NTU games satisfying these
12
properties? The answer is no. It means that some nice things disappear
during the process of extension.
Theorem 6 There is no non-empty solution de…ned on GN with jN j
satisfying symmetry, continuity, and the core property.
3
To obtain the result, we …rst propose a sequence of 3-person games
fVn g1
n=1 which converges to V: The solution of these games is single-valued
which can be determined by non-emptiness, symmetry, and the core property. Then we see that the solution is not continuous. We can view the
class of games as n-person games if we put players other than f1; 2; 3g to be
dummy.
Example 7 Let N = f1; 2; 3g : Assume that is a solution satisfying nonemptiness, symmetry, continuity, and the core property.
Consider the game V 2 GN ; given by
V (S) = GS (f(0; 0; 0)g) for S = f1g ; f2g ; f3g, f1; 2g ; and f1; 3g ;
V (S) = GS ( x 2 RS j x (S) 1 ) for S = f2; 3g ; N:
The core of the game V is
C (V ) = fx 2 V (N ) j x (f2; 3g) = 1 and xi
0 for all i 2 N g :
Let be a permutation such that (1) = 1; (2) = 3; and (3) = 2:
Then is a symmetry of V: Hence, (V ) = 0; 21 ; 12 : Indeed, satis…es
non-emptiness, symmetry, and the core property.
Now we are going to construct a sequence of games fVk g1
k=1 such that
Vk ! V as k ! 1:
1
Denote E = 0; 43 ; 14 ; A = (1; 0; 0) ; Bk = 0; 1 k+3
; 0 ; and Ck =
1
0; 0; 1 k+3 ; where k is any positive integer: Let H (A; Bk ; E) be the half
space determined by A; Bk ; and E and contain the origin 0. Similarly, the
set H (A; Ck ; E) is de…ned.
De…ne Tk = H (A; Bk ; E) \ H (A; Ck ; E) : Please see Figure 2.
Take the smooth surface Wk RN de…ned to be: for each k = 1; 2;
;
!
N
(1) Wk \ x 2 R j x (N ) = 1 = AE, the line determined by A and E.
(2) GN (fWk g) is convex;
(3) Wk lies between @Tk+1 and @Tk :
13
Ck
(0,0,1)
E(0,1/4,3/4)
O
(0,1,0)
A(1,0,0)
Bk
De…ne Vk 2 GN as follows:
Vk (N ) = GN (fWk+1 g) and Vk (S) = V (S) for S 2 P
fN g :
We see that Vk ! V as k ! 1 in the topology induced by extended Hausdor¤
metric.
For every positive integer k; it is clear that C (Vk ) =
0; 43 ; 14 ; and
3 1
hence, (Vk ) = C (Vk ) = 0; 4 ; 4
by the assumption that satis…es nonemptiness and the core property.
We derive that (Vk ) 9 (V ) as k ! 1: It violates that is continuous.
5
The -excess function and the -nucleolus
From Theorem 6, we see that some nice properties cannot be kept if we
extend the nucleolus to NTU games. In particular, both existing extensions
of the nucleolus are not u.h.c. A di¤erent idea to de…ne the excess function is
needed if we expect that the extension of the nucleolus derived by it is u.h.c.
The …rst purpose of the section is to propose the -excess function e~
which is the unique real-valued function satisfying three appealing properties:
excess preservation, excess additivity, and excess invariance.
The second purpose of the section is to de…ne the -nucleolus N (V; e~)
for NTU games. It is interesting to see that N (V; e~) satis…es criteria (A)
and (B) to a large extent. More speci…c, N (V; e~) satis…es the coincidence
14
property, symmetry, and u.h.c. It does not satisfy the core property but
satis…es the inner core property given by Shapley and Shubik(1975) instead.
Besides (A1) and (A2) ; we require a game V in the rest of the paper
satisfying:
(A3) there is a x 2 RN such that V (S) V (N ) + x 8 S 2 P,
(A4) V (N ) is convex with smooth boundary,
(A5) @V (N ) is non-leveled, that is, if x; y 2 @V (N ) and
xi yi 8i 2 N; then x = y:
(A3) can be viewed as a weak monotonicity. (A4) is simply to assure that
any boundary element of V (N ) has a unique supporting hyperplane. (A5) is
equivalent to the property that the normal vector x at x 2 @V (N ) satis…es
( x )i > 0 8i 2 N:
x is interpreted to be “the comparison weights”for the utilities of players.
We assume that it is normalized; that is, ( x )1 = 1: If V corresponds to a TU
game, then x = eN = (1; 1; :::; 1) 2 RN for every x 2 @V (N ) : We denote
0
0
the class of games satisfying (A1) - (A5) to be GN
and GN
GN :
N
Let ! 2 R++ : If x 2 I (V ) ; then ! x 2 I(! V ): We see that x ! 1 is
an normal vector of ! V at ! x 2 I(! V ): Hence, the normalized normal
vector ! x at ! x is ! 1 ( x ! 1 ): If we take ! = x ; then the normalized
normal vector x x of x V at x x is x x = ( x )1 ( x x 1 ) = eN : We
obtain that ( x V ) + V0 corresponds to a TU game by (A3).
Note that the games considered in Example 7 satis…es (A3) - (A5) ; too
0
:
Hence, Theorem 6 still holds in the context of GN
5.1
The - excess function
In this subsection, we de…ne the -excess function e~; and then, axiomatize
0
it: Given a game V 2 GN
and x 2 I (V ) ; we de…ne v x 2 GN by, 8S 2 P,
v
x
(S) = maxf
x
z : z 2 V (S)g:
The -excess function e~ is de…ned from I (V ) RN
0
given x 2 I (V ) ; y 2 RN ; S 2 P, and V 2 GN
;
e~(x; y; S; V ) = v x (S)
x
P
(5)
0
GN
to R as follows:
yS .
v x (S) is well-de…ned because V (S) satis…es (A2) ; (A3) ; and (A4) : It is
interpreted to be the worth of the coalition S 2 P of the game V with respect
15
to the comparison weights x : e~(x; y; S; V ) is to measure “dissatisfaction”of
coalition S at y with respect to the comparison weights x .
If V corresponds to the game v 2 GN ; then, for every x 2 I (V ) ;
e~(x; y; S; V ) = v x (S)
x
yS = v(S)
y(S) = e (y; S; v) :
That is, e~ is the excess for TU games if only games which correspond to TU
games are concerned.
Next, we are going to axiomatize the -excess function e~: Let E~ be a real0
0
valued function de…ned on I (V ) RN P GN
, x 2 I (V ) ; and V 2 GN
:
Excess Preservation(EP): If V corresponds to v 2 GN ; then
~ y; S; V ) = e(y; S; v) 8x 2 I (V ) 8y 2 RN and 8S 2 P:
E(x;
0
Excess Additivity(EA): If V + V0 2 GN
and x 2 I(V + V0 ), then
~ y + 0; S; V + V0 ) = E(x;
~ y; S; V ) + E(0;
~ 0; S; V0 ) 8y 2 RN and 8S 2 P.
E(x;
N
, then
Excess Invariance(EI): If ! = (! 1 ; ! 2 ; :::; ! n ) 2 R++
~
~ y; S; V ) 8y 2 RN and 8S 2 P:
E(!
x; ! y; S; ! V ) = ! 1 E(x;
(6)
The interpretation of EP is that the value of the excess e for TU games
is preserved, that is, the values of E~ and e are the same if V corresponds to a
TU game v. EA says that: Additivity of E~ obtains if one game is restricted
to be the zero game V0 and the payo¤ vector 0. EI says that: If the payo¤s
in V are in utilities, the value of E~ will not be a¤ected if we employ di¤erent
utility functions to represent the same real outcome.
0
x 2 I(! V ): We know
; x 2 I(V ); and ! 2 RN
Given V 2 GN
++ ; !
1
that ( x ! ) is an normal vector of ! V at ! x: So c ( x ! 1 ) is an
normal vector for every c > 0: Since the normalized normal vector is taken
to calculate e~(! x; ! y; S; ! V ); this is why we have ! 1 in (6) : We call
! 1 the normalized f actor:
Lemma 8 The -excess function e~ satis…es EP, EA, and EI.
Proof. We …rst show e~ satisfying EP. Let V and v correspond to each other:
If x 2 I (V ) ; y 2 RN ; and S 2 P, then x = eN and
e~(x; y; S; V ) =
=
=
=
=
v x (S) ( x y)(S)
maxf( x )S z : z 2 V (S)g ( x y) (S)
maxf(eN )S z : z 2 V (S)g y (S)
maxfz(S) : z 2 V (S)g y(S)
e(y; S; v).
16
0
Next, we show e~ satisfying EA. Assume that V + V0 2 GN
and x 2
N
I(V + V0 ). Then V + V0 is a TU game and x = e . Given y 2 RN and
S 2 P, by EP and x = eN ; we have
e~(x; y; S; V + V0 ) =
=
=
=
=
e(y; S; V + V0 )
maxfz(S) : z 2 (V + V0 )(S)g y(S)
maxfz(S) : z 2 V (S)g y(S)
maxf( x )S z : z 2 V (S)g ( x y)(S)
e~(x; y; V; S).
N
It remains to show that e~ satis…es EI. Let ! = (! 1 ; ! 2 ; :::; ! n ) 2 R++
and
1
x 2 I (V ) : Recall that ! x = ! 1 ( x ! ). Then
=
=
=
=
e~(! x; ! y; S; ! V )
maxf( ! x )S z : z 2 (! V )(S)g ( ! x (! y)) (S)
maxf(! 1 ( x ! 1 ))S (! u) : u 2 V (S)g (! 1 ( x ! 1 )) (! y)) (S)
! 1 [maxf( x )S u) : u 2 V (S)g ( x y) (S)]
! 1 e~(x; y; V; S).
Lemma 9 If E~ satis…es EP, EA, and EI, then E~ = e~:
0
and ( x x)+0 2 I( x V +V0 ):
Proof. Let x 2 I (V ) : Then ( x V )+V0 2 GN
N
Given S 2 P and y 2 R ; by EA, EP, e(0; S; v0 ) = 0; EI, and the
assumption that ( x )1 = 1,
~ x x; x y; S; ( x
E(
~ x x; x y; S; x
= E(
~ x x; x y; S; x
= E(
~ x x; x y; S; x
= E(
~ y; S; V )
= ( x )1 E(x;
~ y; S; V ):
= E(x;
V ) + V0 )
~ 0; S; V0 )
V ) + E(0;
V ) + e(0; S; v0 )
V)
Given S 2 P and y 2 RN ; by the fact that (
17
(7)
x
V ) + V0 is a TU game
and EP, we have
=
=
=
=
=
=
~ x x; x y; S; ( x V ) + V0 )
E(
e( x y; S; ( x V ) + V0 )
maxfz(S) : z 2 (( x V ) + V0 )(S)g ( x y) (S)
maxfz(S) : z 2 ( x V )(S)g ( x y) (S)
maxf( x u) (S) : u 2 V (S)g ( x y) (S)
maxf( x )S u : u 2 V (S)g ( x y) (S)
e~(x; y; V; S).
(8)
From (7) and (8) ; we derive the desired result.
Combining previous two lemmas, we have the following:
Theorem 10 e~ is the unique real-valued function de…ned on I (V )
0
satisfying EP, EA, and EI.
P GN
RN
The last part of the subsection is to explain why one more factor I (V ) is
added in the domain of the -excess function e~: Let E be an excess function
0
, we rewrite EP, EA
for NTU games. Since E is de…ned on RN P GN
and EI to be ( ) ; ( ) ; and ( ) ; respectively, in the following. Then we show
that E can be de…ned on hyperplane games only if E satis…es ( ) ; ( ) ; and
0
:
( ) : Let V 2 GN
( ) If V corresponds to v 2 GN ; then
E(y; S; V ) = e(y; S; v) 8y 2 RN and 8S 2 P:
0
( ) If V + V0 2 GN
, then
E(y + 0; S; V + V0 ) = E(y; S; V ) + E(0; S; V0 ) 8y 2 RN and 8S 2 P:
( ) If ! = (! 1 ; ! 2 ; :::; ! n ) 2 RN
++ , then
E(y; S; V ) = ! 1 E(! y; S; ! V ) 8y 2 RN and 8S 2 P:
Note that ! 1 in ( ) is the normalized factor:
0
Theorem 11 Let GN GN
: If E is a function de…ned on RN P
satis…es ( ) ; ( ) ; and ( ), then GN consists of hyperplane games.
18
GN and
Proof. Let V 2 GN and x 2 I (V ) : We see that x x 2 I( x V + V0 ) and
( x V ) + V0 is a TU game.
For S 2 P and an arbitrary y 2 RN ; we have, by ( ) ; the facts that
( x )1 = 1 and E(0; S; V0 ) = e(0; S; v0 ) = 0; ( ) ; and ( ) ;
E(y; S; V ) =
=
=
=
=
=
( x )1 E( x y; S; x V )
E( x y; S; x V ) + E(0; S; V0 )
E( x y; S; ( x V ) + V0 )
e( x y; S; ( x V ) + V0 )
maxf( x )S z : z 2 (( x V ) + V0 )(S)g
v x (S) ( x y)(S):
Since equality holds for every x 2 I (V ) and arbitrary y;
constant. We derive that V is a hyperplane game.
5.2
(
x
x
y) (S)
must be a
The -nucleolus
After having the -excess function e~; we de…ne N (V; e~) ; the nucleolus for
0
; x 2 I (V ),
NTU game V derived by e~; as what we did in (2) : Given V 2 GN
N
and y 2 R ; let
~ (x; y; V; e~)) = (~
e(x; y; S1 ; V );
where Sk 2 P for k = 1; 2;
k = 1; 2; ; 2n 2: De…ne
; 2n
; e~(x; y;S2n 1 ; V )) ;
1 and e~(x; y; Sk ; V )
N (V; e~) = f x 2 IR(V (N )) : ~ (x; x; V; e~)
e~(x; y; Sk+1 ; V ) for
(lex) ~ (x; y; V; e~) 8y 2 IR (V (N ))g:
However, N (V; e~) is not u.h.c. The following example illustrates this.
0
Example 12 Let N = f1; 2; 3g : Let the game V 2 GN
be de…ned by
V (fig) = Gfig (f(0; 0; 0)g) for i 2 N;
V (S) = fx 2 RS : x(S) 0g for S = f1; 2g; f1; 3g;
V (S) = fx 2 RS : x(S) 1g for S = f2; 3g ; N .
Given an arbitrary x 2 I (V ) ; we have
v
x
(S) = 0 if S 2 P
ff2; 3g ; N g ; and v
19
x
(S) = 1 if S = f2; 3g ; N:
C(0,0,1)
Ak
(0,1,0)
O
Bk
(1,0,0)
Then we see that N (V; e~) = f(0; 12 ; 12 )g:
1
Denote A = (1; 0; 0); B = (0; 1; 0); C = (0; 0; 1); Ak = (1 k+3
; 0; 0); and
1
Bk = (0; 1 k+3 ; 0); where k is any positive integer.
Let W (Ak ; Bk ; C) be a convex smooth surface passing Ak , Bk , and C and
satisfy:
(i) @V (N ) = fx 2 RN : x(N ) = 1g is the supporting hyperplane of
W (Ak ; Bk ; C) at C;
(ii) I (V ) \ W (Ak ; Bk ; C) = fCg ; and
(ii) GN (W (Ak ; Bk ; C)) ! V (N ) as k ! 1 in the topology induced by
the extended Hausdor¤ metric.
De…ne the game Vk to be; where k = 1; 2; :::;
Vk (N ) = GN (W (Ak ; Bk ; C)) and Vk (S) = V (S); otherwise.
Please see Figure 3.
0
Then Vk 2 GN
and Vk ! V as k ! 1 in the topology induced by the
extended Hausdor¤ metric.
For arbitrary positive integer k; fCg = N (Vk ; e~) : Indeed, e~(C; C; Vk ; S)
0 8S 2 P and e~(C; x; Vk ; f2; 3g) > 0 if x 2 IR(Vk (N )) fCg:
It implies that N (V; e~) is not u.h.c. because C 6= f(0; 12 ; 21 )g = N (V; e~) :
Besides N (V; e~), we have studied u.h.c. of the other two possible extensions of the nucleolus for NTU games in section 3. All three fail to satisfy the
property. To have an extension of the nucleolus satisfy u.h.c., we are going
to de…ne the -nucleolus N (V; e~) by modifying the idea of (2) slightly.
20
0
0
Given V 2 GN
and x 2 I (V ) ; we denote Vx 2 GN
to be
Vx (S) =
V (S) ;
fy 2 RN j
x
y
if S 2 P fN g ;
v x (N )g; if S = N:
We see that Vx is a quasi hyperplane game and IR (V (N ))
The -nucleolus N (V; e~) of V is de…ned to be
N (V; e~) = fx 2 I (V ) : ~ (x; x; V; e~)
IR(Vx (N )):
(lex) ~ (x; y; V; e~) 8y 2 IR(Vx (N ))g:
The di¤erence between the nucleolus N (V; e~) and the -nucleolus N (V; e~)
is that the former chooses the smallest element to minimize the “total dissatisfaction” among IR (V (N )) and the latter chooses the smallest element
among the bigger set IR(Vx (N )).
We know that Shapley(1969) and Otten and Peters(2002) each proposed
an existence result of the NTU value by taking advantage of the fact that
the Shapley value is continuous. Their ideas can be employed to study the
existence of N (V; e~) because the nucleolus for TU games satis…es continuity,
too.
0
. The inner core de…ned by Shapley and Shubik(1975) is the
Let V 2 GN
following:
S
IC (V ) =
fy 2 IR (V (N )) : x yS v x (S) 8S 2 Pg (9)
x2I(V )
=
S
x2I(V )
fy 2 IR (V (N )) : e~(x; y; V; S)
0 8S 2 Pg .
0
As for the existence of the inner core, please see Qin(1994). Given V 2 GN
and IC (V ) 6= ;; we say that the solution satis…es the inner core property
if (V ) IC (V ) :
0
Given ! 2 RN
++ and V 2 GN ; we denote (! V ) (S) = f! y : y 2 V (S)g ;
0
where S 2 P: Then ! V 2 GN
:
0
Theorem 13 Let V 2 GN
and x 2 I (V ) :
(i) x 2 N (V; e~) if and only if f x xg = N (v x ; e) :
0
(ii) N (V; e~) is u.h.c. in GN
:
(iii) N (V; e~) satis…es the coincidence property, symmetry, and inner core
property.
21
Proof. If x 2 I(V );
(
x
Vx ) (N ) =
x
=
x
y 2 RN :
y2R :(
z2R
=
x
N
N
y
v x (N )
y) (N )
x
: z (N )
v x (N )
v x (N ) :
For i 2 N;
v
x
(fig) = maxf
x
z : z 2 V (fig)g = maxfyi : y 2 (
x
V ) (fig)g:
From the previous two observations, we see that
I (v x ) = I(
x
(10)
Vx ):
We …rst prove "only if" part of (i) : Given y 2 Vx (N ) and S 2 P; we have
e~(x; y; S; V ) = v x (S)
x
yS = v x (S)
(
x
y)(S) = e (
x
y; S; v x ) ;
and hence,
~ (x; y; V; e~)) = (~
e(x; y; S1 ; V );
; e~(x; y;S2n 1 ; V ))
= (e( x y; S1 ; v x );
; e( x y; S2n 1 ; v x ))
= ( x y; v x ; e) :
(11)
Hence, x 2 N (V; e~) if and only if x 2 I(V ) and ~(x; x; V; e~)
(lex)
~(x; y; V; e~) 8y 2 IR(Vx (N )). It implies that ( x x; v x ; e) (lex) ( x
y; v x ; e) 8 x y 2 I(v x ) by (11) and (10) : We obtain that f x xg =
N (v x ; e) :
As for the "if" part of (i) ; it can be derived by similar arguments.
0
Now we prove (ii) : Let fVk g1
k=1 be a sequence of games in GN and Vk ! V
as k ! 1 in the topology induced by the extended Hausdor¤ metric.
Let xk 2 N (Vk ; e~) 8 k and xk ! x 2 I (V ) as k ! 1: Then xk ! x
and xk xk ! x x as k ! 1 in the topology induced by the Euclidean
distance because @Vk (N ) is smooth for every k by (A4). Due to (5) ; we see
that (vk ) xk ! v x as k ! 1.
Since f xk xk g = N (vk ) xk ; e 8k; xk xk ! x x = N (v x ; e) as
k ! 1: Indeed, the nucleolus is continuous on TU games. We derive that
x 2 N (V; e~) by (i).
It remains to prove (iii) : Let V correspond to the TU game v. Then x =
N
e 8 x 2 @V (N ) and v x = v: It implies that N (V; e~) = N (v; e) by (i) ;
22
and hence, N (V; e~) satis…es the coincidence property. It is straightforward
to derive that N (V; e~) satis…es symmetry and inner core property. We omit
the proofs. (iii) is obtained.
The following example demonstrates that N (V; e~) does not satisfy the
core property.
0
Example 14 Let N = f1; 2; 3g : Consider the game V 2 GN
given by
V (fig) = Gfig (f(0; 0; 0)g), where i 2 N ,
V (S) = fx 2 RS : x(S) 2g; where S = f1; 2g ; f2; 3g ;
V (f1; 3g) = Gf1;3g (f(1; 0; 2)g), and V (N ) = x 2 RN j x (N )
3 :
We see that C(V ) = fx 2 RN : x is in the line segment determined by (1; 1; 1)
and (1; 2; 0)g.
= (1; 1; 1) is the normalized normal vector at every x 2 I (V ). Hence,
we have
v (fig) = 0 for i = 1; 2; 3,
v (f1; 2g) = v (f2; 3g) = 2, and v (f1; 3g) = v (N ) = 3.
Then N (v ; e) =
4 1 4
; ;
3 3 3
62 C (V ).
Remark 15 Property I and property II are two interesting properties because
each one can be used to characterize and to prove continuity of the nucleolus
for TU games, please see Kohlberg(1971) : From (i) of Theorem 13, we see
that the -nucleolus also shares these two properties in the following sense.
0
; x 2 N (V; e~) if and only if the coalition array b1 ;
; bp that
Given V 2 GN
belongs to ( x x; v x ) satis…es property I and property II.
6
Shapley procedure
We have observed that the -nucleolus satis…es many appealing properties.
However, to claim a solution is game-theoretically sound, one usually tries
to provide an axiomatization of it. Fortunately, it is not di¢ cult to propose
one for the -nucleolus.
0
Let V 2 GN
: Given a solution for TU games '; if ' is a solution for NTU
games and
' (V ) = fx 2 I (V ) : x x 2 ' (v x )g ;
(12)
23
we say that the solution ' is derived from ' via Shapley procedure. It is
known that ' is the NTU value if ' is the Shapley value. From (9) ; we see
that ' is the inner core if ' is the core. Although the -nucleolus is derived
by a di¤erent way, it turns out to be the nucleolus version derived by the
procedure, please see (i) of Theorem 13:
(12) provides an interesting relation between ' and ': It makes ' be able
to function like a solution for TU games. Aumann(1985) axiomatized the
NTU value by taking advantage of the relation. The idea to show that the
-nucleolus is u.h.c. depends on it, too. We see that any pair of solutions, a
solution for TU games and its corresponding solution for NTU games derived
via Shapley procedure, share the relation. The purpose of the section is
to illustrate that the ideas used to obtain the two facts can be applied to
each solution for NTU games derived via the procedure. In particular, an
axiomatization of the -nucleolus is obtained.
Denote G'
GN to be the set of games with ' (v) 6= ; 8v 2 G' and
0
G'
GN to be the set of games with ' (V ) 6= ; 8V 2 G' : We …rst study
u.h.c. of the solution ':
Theorem 16 Let ' be a solution de…ned on G' : Then ' is u.h.c. on G' if
' is u.h.c.
We omit the proof because it can be obtained by following the same idea
that we show that N (V; e~) is u.h.c. in (iii) of Theorem 13.
Next, we are going to study axiomatization of ': Here we assume that '
is de…ned on G' and satis…es:
( ) ' (v0 ) = 0; where v0 is the zero game;
( ) ' is covariant under S-equivalence, that is; given u; v 2 G' ; r > 0;
2 RN ; and v (S) = ru (S) + (S) for all S N; then ' (v) = r' (u) + ;
and
( ) ' (v) I (v) :
Note that almost all major solutions for TU games satisfy ( ) ; ( ) ; and ( ).
Let be a solution de…ned on G' and satisfy the following axioms: for all
games U; V; W 2 G' ;
Axiom 1 Non-emptiness(N E) : (V ) 6= ;:
Axiom 2 E¢ ciency(EF F ): (V ) @V (N ) :
Axiom 3 Restricted Additivity(RA) : If U = V + V0 ; then
(U )
( (V ) + (V0 )) \ @U (N ) :
24
Axiom 4 Scale Covariance(SC) : (
V) =
(V ) :
Axiom 5 Independence of Irrelevant Alternatives : If V (N ) W (N ) and
V (S) = W (S) for all S 2 P fN g ; then (V )
(W ) \ V (N ) :
Axiom 6 Coincidence Property(CP ) : Let v 2 GN and V 2 G' correspond
to each other. Then (V ) = ' (v) :
For interpretations of Axioms 1, 2, 4, and 5, please see Aumann(1985) :
Restricted additivity means that if x is in the solution (V ) and it does
happen to be e¢ cient in V + V0 ; then it is also in the solution (V + V0 ) =
(U ) : That is, additivity obtains if one of the games is a zero game. If one
claims to extend a solution for TU games to NTU cases, it is reasonable to
require that it coincides with its original version on TU games. This is what
Axiom 6 interprets.
Remark 17 Most of axioms in the axiomatic system used to axiomatize the
NTU value in Aumann(1985) are kept except the following three: Unanimity,
Conditional Additivity, and Closure Invariance.
(1) Unanimity is replaced by Coincidence Property.
(2) Conditional Additivity is replaced by Restricted Additivity.
(3) Closure Invariance is deleted because of (A1).
Lemma 18 ' satis…es N E; EF F; RA; SC; IIA;and CP on G' :
Proof. ' satis…es N E because it is de…ned on G' :
' satis…es EF F because of (12) :
' satis…es CP because of (12) :
' satis…es IIA and SC: These can be obtained straightforwardly. Note
that we need ( ) to have SC:
It remains to show that ' satis…es RA: We know that f0g = ' (V0 ) by
( ) and CP: Let y 2 ' (V ) and y 2 @U (N ) : We wish to have y 2 ' (U ) : Let
y be the normalized normal vector of U at y: Then y = (1; 1; :::; 1) because
U corresponds to a TU game. Since v + v0 = v and ' (v0 ) = 0 by ( ) ; we
have '(v + v0 ) = '(v) + '(v0 ): Then RA can be derived by following what
Aumann showed the NTU value satis…es Conditional Additivity.
Lemma 19 ' (V ) = (V ) 8V 2 G'
The Lemma can be derived by following the proofs of Lemmas 8.6 and
8.7 of Aumann(1985) step by step. We omit the proof.
Note that Lemma 8.1 of Aumann(1985) is exactly Axiom 6 Coincidence
Property. Then we have the following result from previous two lemmas.
25
Theorem 20 ' is the unique solution satisfying N E; EF F; RA; SC; IIA,
and CP .
From Theorem 16 and Theorem 20, we see that those solutions for NTU
games derived from Shapley procedure satisfy the coincidence property and
preserve many nice properties of their corresponding solutions for TU games,
for instance, u.h.c., e¢ ciency, scale covariance, etc. Furthermore, every such
a solution for NTU games is axiomatized. Therefore, if one tries to extend a
solution for TU games to NTU cases, Shapley procedure is recommended to
do it.
References
[1] Aliprantis CD and Border KC(2006) In…nite Dimensional Analysis: A
Hitchhiker’s Guide. Springer.
[2] Aumann R(1985) An axiomatization of the non-transferable utility
value. Econometrica 53: 599-612.
[3] Dugundji J(1966) Topology. Allyn and Bacon.
[4] Kalai E (1975) Excess function for cooperative game without side payments. SIAM J Appl Math 29: 60-71.
[5] Klauke S(2002) NTU prenucleoli. working paper 329, University of Bielefeld, Germany.
[6] Kohlberg E (1971) On the nucleolus of a characteristic function game.
SIAM J Appl Math 20: 62-66.
[7] Kaneko M and Wooders M(2004) Utility theories in cooperative games.
Handbook of Utility Theory II Eds. S. Barbera et al. Kluwer Academic
Pub.
[8] Nakayama M (1983) A note on a generalization of the nucleolus to games
without side payments. Int J Game Theory 12: 115-122.
[9] Otten GJ and Peters H(2002) The Shapley transfer procedure for NTU
games. Chapters in game theory: in honor of Stef Tijs, Kluwer Academic
Pub.
26
[10] Qin C.-Z. (1994) The Inner Core of an n-Person Game. Games and
Economic Behavior 6: 431-444.
[11] Rockafellar, RT (1970) Convex Analysis. Princeton University Press.
[12] Schmeidler DN (1969) The nucleolus of a characteristic function game
form. SIAM J Appl Math 17: 1163-1170.
[13] Shapley LS (1969) Utility comparison and the theory of games. In La
Decision, Paris: Editions du CNRS, 251-263.
[14] Shapley LS and Shubik M(1975) Competitive outcomes in the cores of
market games. Int J Game Theory 4: 229-237.
27